Let $h(x)=x^3+7e^x$. $h'(x)=$
Recall that ${\dfrac{d}{dx}[e^x]=e^x}$ and ${\dfrac{d}{dx}[x^n]=nx^{n-1}}$. $\begin{aligned} h'(x)&=\dfrac{d}{dx}[x^3+7e^x] \\\\ &={\dfrac{d}{dx}[x^3]}+7{\dfrac{d}{dx}[e^x]} \\\\ &={3x^2}+7{e^x} \end{aligned}$ In conclusion, $h'(x)=3x^2+7e^x$